Question

Can any ring be embedded as a subring of a left semisimple ring?

Answer 1

The answer is "no": for instance, $A = R1 \times R2 \times \cdots$ (where $Ri$ are nonzero rings) cannot be embedded as a subring of a (left) semisimple ring. Indeed, if $A$ is a subring of a ring $R$, then $R$ will have nonzero idempotents $e1, e2, \ldots$ with $ei*ej = 0$ for $i <> j$. But then $R \supseteq Re1 \circ Re2 \circ \cdots$, and this implies that $R$ is not left noetherian (let alone left semisimple).

Similarly, if $k$ is any nonzero ring, $A = k[x1, x2, \ldots]$ with the relations $xi * xj = 0$ (for all $i, j$) cannot be embedded in a left semisimple ring. Indeed, if $A$ is a subring of a ring $R$, then $Rx1 \subset Rx1 + Rx2 \subset \cdots$ (by an easy proof), and again $R$ is not left noetherian (let alone left semisimple).